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Find the equation of a line parallel to 2x + 3y = - 15 that passes through the point (3, 4)

User Crawfish
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Given:

a.) Parallel to a line: 2x + 3y = - 15

b.) Passes through the point (3, 4)

Let's find the equation of a line parallel to 2x + 3y = -15.

Note: Since the line that we are looking at is parallel to 2x + 3y = -15, they should have the same slope.

Step 1: Let's determine the slope of the line based on its slope-intercept form: y = mx + b.


\begin{gathered} \text{ 2x + 3y = -15} \\ \text{ 3y = -15 - 2x} \\ \text{ }\frac{\text{3y}}{\text{ 3}}\text{ = }\frac{\text{-2x - 15}}{3} \\ \text{ y = -}\frac{2}{\text{ 3}}x\text{ - 15} \end{gathered}

Therefore, the slope of the line (m) is -2/3.

Step 2: Using the slope = -2/3 and x,y = 3,4. Plug it in y = mx + b to find the y-intercept of the parallel line.


\begin{gathered} \text{ y = mx + b} \\ \text{ 4 = (-}(2)/(3))(3)\text{ + b} \\ \text{ 4 = -2 + b} \\ \text{ b = 4 + 2} \\ \text{ b = 6} \end{gathered}

Step 3: Let's now complete the equation. Substitute the slope (m) = -2/3 and b = 6 in y = mx + b.


\begin{gathered} \text{ y = mx + b} \\ \text{ y = (-}(2)/(3))x\text{ + (6)} \\ \text{ y = -}(2)/(3)x\text{ + 6} \end{gathered}

Therefore, the equation of the line parallel to 2x + 3y = -15 is:


\text{ y = -}(2)/(3)x\text{ + 6}

User AlexeyVMP
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